Answer:
The gauge pressure is [tex]1.8\times10^{5}\ N/m^2[/tex]
Explanation:
Given that,
Gauge pressure of car tires [tex]P_{1}=2.40\times10^{5}\ N/m^2[/tex]
Temperature [tex]T_{1}=35.0^{\circ}C = 35.0+273=308 K[/tex]
Dropped temperature [tex]T_{2}= -42.0^{\circ}C=273-42=231 K[/tex]
We need to calculate the gauge pressure P₂
Using relation pressure and temperature
[tex]\dfrac{P_{1}}{T_{1}}=\dfrac{P_{2}}{T_{2}}[/tex]
Put the value into the formula
[tex]\dfrac{2.40\times10^{5}}{308}=\dfrac{P_{2}}{231}[/tex]
[tex]P_{2}=\dfrac{2.40\times10^{5}\times231}{308}[/tex]
[tex]P_{2}=180000 = 1.8\times10^{5}\ N/m^2[/tex]
Hence, The gauge pressure is [tex]1.8\times10^{5}\ N/m^2[/tex]
